1. The photoelectric effect
Contents:
Einstein’s proposed experiment
Solving photoelectric problems
Example 1 | Example 2
Whiteboard
Photon vs wave theory

2. Light
Waves Photons
Wavelength Changes
Energy per photon changes
E = hf
Radio waves vs X-rays
Color
small  = blue
big  = red
Amplitude Changes
# of photons changes
many = bright
few = dim
CCD Devices, High speed film
small = dim
Brightness
big = bright

3. Photo Electric Effect Reverse the voltage
How it works
Einstein’s idea
Photon Energy = Work + Kinetic Energy
hf =  Emax
hf = hfo eV
 - Work function (Depends on material)
fo - Lowest frequency that ejects
e - Electron charge
V - The uh stopping potential + - V
Reverse the voltage
Ex. 1:  = 3.25 eV, V = 7.35 V,  = for light?
Ex. 2:  = 70.9 nm,  = 5.10 eV, V = for electrons?

4. Photon Energy = Work + Kinetic Energy
hf =  Emax
hf = hfo eV
 - Work function (Depends on material)
fo - Lowest frequency that ejects
e - Electron charge
V - The uh stopping potential
Example 1: A certain metal has a work function of 3.25 eV. When light of an unknown wavelength strikes it, the electrons have a stopping potential of 7.35 V. What is the wavelength of the light?
Photon energy = Work + Kinetic Energy
Photon Energy = 3.25 eV eV = eV
Convert to J: (10.60 eV)(1.602E-19 J/eV) = E-18 J
E = hf = hc/ ,  = hc/E = (6.626E-34 Js)(3.00E8 m/s)/( E-18 J)
= E-07 m = 117 nm

5. Photon Energy = Work + Kinetic Energy
hf =  Emax
hf = hfo eV
 - Work function (Depends on material)
fo - Lowest frequency that ejects
e - Electron charge
V - The uh stopping potential
Example 2: 70.9 nm light strikes a metal with a work function of 5.10 eV. What is the maximum kinetic energy of the ejected photons in eV? What is the stopping potential?
E = hf = hc/ = (6.626E-34 Js)(3.00E8 m/s)/(70.9E-9)
= E-18 J
Convert to eV: ( E-18 J)/(1.602E-19 J/eV) = 17.5 eV
Photon Energy = Work + Kinetic Energy
17.5 eV = Kinetic Energy
Kinetic Energy = 12.4 eV, so 12.4 V would stop the electrons.

6. Whiteboards: Photoelectric effect 1 | 2 | 3 | 4 Ag (silver) 4.26
Au (gold)
5.1
Cs (cesium)
2.14
Cu (copper)
4.65
Li (lithium)
2.9
Pb (lead)
4.25
Sn (tin)
4.42
Chromium
4.6
Nickel
Metal
Work Function
Whiteboards:
Photoelectric effect
1 | 2 | 3 | 4

7. Photons of a certain energy strike a metal with a work function of 2
Photons of a certain energy strike a metal with a work function of 2.15 eV. The ejected electrons have a kinetic energy of 3.85 eV. (A stopping potential of 3.85 V) What is the energy of the incoming photons in eV?
Photon energy = Work function + Kinetic energy of electron
Photon energy = 2.15 eV eV = 6.00 eV
6.00 eV

8. Another metal has a work function of 3. 46 eV
Another metal has a work function of 3.46 eV. What is the wavelength of light that ejects electrons with a stopping potential of 5.00 V? (2)
E = hf = hc/,
Photon energy = Work function + Kinetic energy of electron
Photon energy = 3.46 eV eV = 8.46 eV
E = (8.46 eV)(1.602 x J/eV) = x J
 = hc/E = (6.626 x Js)(3.00 x 108 m/s)/( x J)
 = x m = 147 nm
147 nm

9. 112 nm light strikes a metal with a work function of 4. 41 eV
112 nm light strikes a metal with a work function of 4.41 eV. What is the stopping potential of the ejected electrons? (2)
E = hf = hc/, 1 eV = x J
Photon energy = Work function + Kinetic energy of electron
E = hf = hc/ = (6.626 x Js)(3.00 x 108 m/s)/(112 x 10-9 m)
E = x J
E = ( x J)/(1.602 x J/eV) = eV
eV = 4.41 eV eVs
eV eV = eV = eVs
Vs = 6.67 V
6.67 V

10. 256 nm light strikes a metal and the ejected electrons have a stopping potential of 1.15 V. What is the work function of the metal in eV? (2)
E = hf = hc/, 1 eV = x J
Photon energy = Work function + Kinetic energy of electron
E = hf = hc/ = (6.626 x Js)(3.00 x 108 m/s)/(256 x 10-9 m)
E = x J
E = ( x J)/(1.602 x J/eV) = eV
4.847 eV = Work function eV
eV eV = 3.70 eV
3.70 eV

12. Einstein’s Photon theory predicts:
Photon energy = work function + Kinetic energy of electron
hf =  + Ekmax
Ekmax = hf - 
Photon Theory predicts:
Ekmax rises with frequency
Intensity of light ejects more
Wave Theory predicts:
Ekmax rises with Amplitude (Intensity)
Frequency should not matter
Survey says
Millikan does the experiment
1915 conclusion
1930 conclusion